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April 28th, 2015, 01:25 PM  #1 
Newbie Joined: Apr 2015 From: England Posts: 6 Thanks: 0  Please help me to understand
I have recently stumbled across this following question: The answers are the following: I would like to know how to work out the answers as I got them wrong. i) I understand why it would be xy, but I don't understand why if it is xy the number would be 2^p+s and not 4^p+s. ii) I completely don't understand this. iii) Completely confused with this one as well. It would be great if someone could help and say how those answers came about. Thanks a lot. Last edited by skipjack; April 28th, 2015 at 03:57 PM. 
April 28th, 2015, 01:33 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,661 Thanks: 965 Math Focus: Elementary mathematics and beyond 
i) $\displaystyle 2^{p+s}=(2^p)(2^s)=xy$ ii) $\displaystyle 2^{2p}=(2^p)^2=x^2$ iii) $\displaystyle 2^{p1}=(2^p)(2^{1})=2^p\cdot\frac12=x\cdot\frac12=\frac12x$ 
April 28th, 2015, 01:50 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,091 Thanks: 2360 Math Focus: Mainly analysis and algebra 
Recall that (for any positive integers $p$ and $s$) $$2^p = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} \\ 2^s = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$s$ terms}} $$ So when we multiply we get $$2^p \times 2^s = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} \times \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$s$ terms}} = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p+s$ terms}} = 2^{p+s}$$ Similarly, $$\left(2^p\right)^2 = 2^p \times 2^p = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} \times \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p+p$ terms}} = 2^{p+p} = 2^{2p}$$ Finally, $$2^{p1} \times 2 = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p1$ terms}} \times 2 = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} = 2^p$$ so we can divide by 2 to get $$2^{p1} = \frac12 2^p$$ These results generalise to any $p$ and $s$. 
April 28th, 2015, 02:03 PM  #4  
Newbie Joined: Apr 2015 From: England Posts: 6 Thanks: 0  Quote:
Quote:
thank you very much (: One more thing, as you said you explained the p+s terms come together to form 2^p+s But how come when you do a multiplication such as 9^3 x 9^3 = It equals 81^6 and not 9^6 Last edited by nautudent; April 28th, 2015 at 02:07 PM.  
April 28th, 2015, 02:26 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,091 Thanks: 2360 Math Focus: Mainly analysis and algebra  
May 2nd, 2015, 05:08 AM  #6 
Newbie Joined: Apr 2015 From: England Posts: 6 Thanks: 0  
May 2nd, 2015, 05:34 AM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,091 Thanks: 2360 Math Focus: Mainly analysis and algebra 
$$\begin{aligned}4pq \times 6p^2q &= 4 \times p \times q \times 6 \times p \times p \times q \\ &= \underbrace{4 \times 6}_{24} \times \underbrace{p \times p \times p}_{p^3} \times \underbrace{q \times q}_{q^2} \\ &= 24p^3q^2\end{aligned}$$

May 2nd, 2015, 05:44 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 18,247 Thanks: 1439 
One can add the exponents when finding the product of powers of the same thing: $p^mp^n = p^{m+n}$. If the exponents are equal, one can find the product as follows: $p^nq^n = (pq)^n$. Hence $9^39^3$ is equal to both $9^6$ and $81^3$. 

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